Self Numbers 分类： POJ 2015-06-12 20:07 14人阅读 评论(0) 收藏

Self Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22101		Accepted: 12429

Description

In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to be n plus the sum of the digits of n. (The d stands for digitadition, a term coined by
Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 +
3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence




33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ...

The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators, 91 and 100. A number with
no generators is a self-number. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.

Input

No input for this problem.

Output

Write a program to output all positive self-numbers less than 10000 in increasing order, one per line.

Sample Input

Sample Output

1
3
5
7
9
20
31
42
53
64
 |
 |       <-- a lot more numbers
 |
9903
9914
9925
9927
9938
9949
9960
9971
9982
9993

就是筛选出10000以内的Self Numbers，类似素数筛

#include <cstdio>
#include <string.h>
#include <cmath>
#include <iostream>
#include <algorithm>
#define WW freopen("output.txt","w",stdout)
using namespace std;
const int Max=10000;
bool vis[Max];
int main()
{
    memset(vis,false,sizeof(vis));
    for(int i=1; i<Max; i++)
    {
        if(!vis[i])
        {
            int ans=i;
            while(ans<Max)
            {
                int ant=ans;
                while(ans)
                {
                    ant+=(ans%10);
                    ans/=10;
                }
                ans=ant;
                if(!vis[ans])
                    vis[ans]=true;
                else
                {
                    break;
                }
            }
        }
    }
    for(int i=1; i<Max; i++)
    {
        if(vis[i])
            continue;
        printf("%d\n",i);
    }
    return 0;
}

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